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Kyle's mother traveled the same 12 mile route as kyle, but she drove her car. During the first third, kyle's mother traveled at an average rate four times greater than kyle's. During the second third of the trip, she traveled at an average rate 4 miles less per hour less than the first third. During her last third of the trip, she traveled downhill at an average rate of 12 miles per hour greater than the second section. How long did it take kyle's mother to travel the same 12 miles by car that kyle traveled on his bike

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Answer:

The time it would take Kyle's mum is given as follows;

[tex]t = \dfrac{3 \cdot v^2+ 2\cdot v - 2}{v \cdot (v - 1) \cdot (v + 2)}[/tex]

Where v = Kyle's average rate

Step-by-step explanation:

The details of Kyle's mother's journey are;

The distance she drove = 12 miles

Her speed during the first third (12/3 = 4 miles), v₁ = 4 × Kyle's speed, v

Her speed in the second third(between 4 and 8 miles), v₂ = v₁ - 4

Her speed in the last third, v₃ = v₂ + 12

∴ v₁ = 4·v

v₂ = v₁ - 4 = 4·v - 4

v₃ = v₂ + 12 = 4·v - 4 + 12 = 4·v + 8

∴ t₁ = (t/3)/4 = t/12 = 4/(4·v) = 1/v

t₂ = 4/(4·v - 4) = 1/(v - 1)

t₃ = 4/(4·v + 8) = 1/(v + 2)

The total time, t = t₁ + t₂ + t₃

∴ t = 1/v + 1/(v - 1) + 1/(v + 2) = [tex]\dfrac{3 \cdot v^2+ 2\cdot v - 2}{v \cdot (v - 1) \cdot (v + 2)}[/tex]

The time it would take Kyle's mum to travel the same 12 miles by car that Kyle traveled on his bike, t = [tex]\dfrac{3 \cdot v^2+ 2\cdot v - 2}{v \cdot (v - 1) \cdot (v + 2)}[/tex]

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